7. The convolution of distributions 27
Proposition 7.4. For any f E there exists C0
∞(Rn)
such that
f in D as ε 0.
Proof: Let βε be a delta-like sequence as in Example 4.4, and, in ad-
dition, β(x) C0
∞(B1).
Let fε(x) = f βε = f(βε(x ·)). By (7.6) we
have:
fε(ϕ) = f(βε1 ϕ),
where βε1(x) = βε(−x). For any k we have
∂k
∂xk
(βε1 ϕ) = βε1
∂kϕ
∂xk

∂kϕ
∂xk
uniformly (see Proposition 1.3). Since supp(βε1 ϕ) BR+1 for ε 1,
where supp ϕ BR (see Proposition 1.2), we have that βε1 ϕ ϕ in D.
Therefore fε(ϕ) = f(βε1 ϕ) f(ϕ).
Example 7.1. For any f D we have f δ = f. Indeed, by (7.9),
(f δ)(ψ) = f(δ ψ). We used that δ(−x) = δ(x). By (7.1), δ ψ =
δ(ψ(x ·)) = ψ(x). Thus (f δ)(ψ) = f(ψ) for any ψ D. Similarly, we
have f
∂kδ
∂xk
=
∂kf
∂xk
.
7.3. Direct product of distributions.
Let f1
D(Rn1
) and f2 D
(Rn2
) and let x = (x , x ), where x
Rn1
, x
Rn2
. For any ϕ(x , x ) C0
∞(Rn1+n2
) the function ϕ1(x ) = f1(ϕ(·,x ))
C0
∞(Rn2
) (cf. the proof of Proposition 7.2). If ϕn(x , x ) ϕ(x , x ) in
D(Rn1+n2
), then ϕ1n(x ) = f1(ϕn(·,x )) ϕ1(x ) in
D(Rn2
). Therefore
f2(ϕ1) is well defined, and we call f2(ϕ1) = f2(f1(ϕ)) the direct product
(f1 × f2) D
(Rn)
of f1 and f2:
(7.12) (f1 × f2)(ϕ) = f2(f1(ϕ(x , x ))), ∀ϕ C0
∞(Rn1+n2
).
Analogously,
(f2 × f1)(ϕ) = f1(f2(ϕ(x , x )))
D(Rn1+n2
).
Note that f1 × f2 = f2 × f1. This is obvious in the case where ψ(x , x ) =
∑N
j=1
ψj1(x )ψj2(x ), with ψj1(x ) C0
∞(Rn1
), ψj2(x ) C0
∞(Rn2
), since
(f1 × f2)(ψ) =
N
j=1
f1(ψj1)f2(ψf2) = (f2 × f1)(ψ).
For any ϕ C0
∞(Rn1+n2
), there is a sequence of C0
∞(Rn1+n2
) functions of the
form
∑N
j=1
ψj1(x )ψj2(x ) converging to ϕ(x , x ) in
D(Rn1+n2
). Therefore
we get that (f1 × f2)(ϕ) = (f2 × f1)(ϕ) for all ϕ
D(Rn1+n2
).
Proposition 7.5. Let x = (x , xn) and let f D
(Rn).
If f(
∂ϕ
∂xn
) = 0 for
any ϕ(x , xn) C0
∞(Rn),
then f = f1 × 1, where f1 D
(Rn−1)
and 1 is the
regular functional in D
(R1)
corresponding to 1.
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